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Piyush Theroem
Piyush Theorem: In a Right-Angled Triangle with sides in A.P Theorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median & altitude at the base is 1/10 Th the sum of other two sides. This Theorem applies in Two Conditions: # The Triangle must be Right-Angled. # Its Sides are in A.P. Series. # Proof with Trigonometry Tan α =AD/DC AD= DC Tan α —————–1 Tan α = AD/DE AD= DE Tan2α —————-2 DC Tan α = DE Tan 2α (DE+EC) Tan α = DE Tan 2α DE Tan α + EC Tan α = DE Tan 2α DE Tan α + EC Tan α = 2 DE Tan α / (1- Tan^2 α ) DE Tan α – DE Tan^3 α + EC Tan α –EC Tan^3 α = 2DE Tan α EC Tan α –EC Tan^3 α– DE Tan^3 α = 2DE Tan α – DE Tan α Tan α (EC – EC Tan^2 α – DE Tan^2 α )= DE Tan α DE Tan^2 α – DE = EC Tan^2 α – EC -DE ( Tan^2 α + 1) = -EC (1 – Tan^2 α) DE (sin^2α /cos^2α + 1) = EC (1- sin^2α /cos^2α) DE (sin^2α+ cos^2α)/cos^2α = EC (cos^2α– sin^2α)/cos^2α DE (sin^2α + cos^2α) = EC(cos^2α –sin^2α) DE (sin^2α + cos^2α) = EC (cos^2α –sin^2α ) where (sin^2α + cos^2α =1) & (cos^2α –sin^2α=cos^2α ) DE= EC cos^2α cosα =a/a+d & sinα= (a-d)/ (a +d) cos^2α = a^2/ (a +d)^2 sin^2α = (a-d) ^2/ (a+ d)^ 2 DE= EC (cos^2α – sin^2α) = EC (a^2 / (a +d)^ 2 – (a-d) ^2/ (a +d) ^2 = EC (a^2 – (a-d) ^2/ (a +d) ^2 = EC (a –a +d) (a+ a-d)/ (a+ d)^ 2 = EC (d)(2a -d)/(a+ d) ^2 = (a +d)/2(d) (2a -d)/ (a +d)^ 2 ————- where EC= (a +d)/2 = (d) (2a -d)/2(a +d) = (d) (8d -d)/2(4d+d) ——————where a= 4d (as per the Theorem) = 7d^2 /2(5d) = 7d /10 = (3d+4d)/10= (AB+AC)/10 2.Proof with Obtuse Triangle Theorem AC^2=EC^2 +AE^2 +2CE.DE where EC = ( a +d) /2,AE=( a +d)/2 a^2 = (a +d/2)^2 + (a+ d/2)^2 + 2(a +d)/2DE = (a +d/2) (a+d+2DE) = (a +d/2) (a+d+2DE) where a=4d 16d^2 = (5d/2) (5d+2DE) 32d/5 = 5d + 2DE 32d/5 – 5d = 2DE 32d -25d/5 = 2DE DE =7d/10 = (3d+4d)/10 = (AB+AC)/10 3.Proof with Acute Triangle Theorem AB^2= AC^2+BC^2 – 2BC.DC (a-d) ^2= a^2 + (a+ d)^ 2 -2(a+ d) (DE+EC) where AB= (a-d), AC=a, BC =( a +d) & EC= (a +d)/2 (a-d)^ 2 – (a +d)^2 = a^2 -2(a +d)(DE+EC) (a- d –a-d) (a -d +a +d) = a^2 -2(a+ d) (2DE+a+d)/2 2(-2d) (2a) = 2a^2 -2(a +d) (2DE+a+d) -8ad – 2a^2 = -2(a +d) (2DE+a+d) -2a (4d +a) = -2(a +d) (2DE+a+d) a (4d + a) = (a +d)(2DE+a+d) 4d (4 d + 4d) = (4d+d) (2DE+4d+d) 4d (8d) = (5d) (2DE+5d) 32d2/5d = (2DE+5d)http://gonitsora.com/theorem-right-angled-triangles/ 32d/5 = (2DE+5d) 32d/5 – 5d = 2DE (32d – 25d)/5 = 2 DE DE = 7d/10 = (3d+4d)/10 = (AB+AC)/10 4. Proof with Co-ordinates Geometry'''1.https://edupediapublications.org/journals/index.php/IJR/article/view/3743/3589 2.http://gonitsora.com/theorem-right-angled-triangles/ Equation of BE Y – 0 =b-0/0-a(X – a) Y = -b/a(X) + b——————- (1) M1 = -b/a For perpendicular M1M2= -1 So M2=a/b Equation of AC Y – 0 = a/b(X-0) Y=a/b(X) —————— (2) Put Y value in equation (1) a/b(X) + b/a(X) =b X (a^2+b^2/a b) = b X = ab^2/ (a^2 + b^2) To get Value of Y, put X value in equation (2) Y = a/b (ab^2/ (a^2+b^2) Y = a^2b/ (a^2+b^2) Here we got co-ordinates of Point C – ab^2/ (a^2 + b^2), a^2b/ (a^2+b^2) and co-ordinates of point d is (a/2, b/2) because d is midpoint. As per the “Theorem” a=z-d, b=z, c = z+ d (z +d) ^2= (z-d) ^2+z^2 from here z=4d so a=3d and b=4d Put value of a & b ab^2/ (a^2 + b^2), a^2b/ (a^2+b^2) & (a/2, b/2) ab^2/ (a^2 + b^2) = 48d/25 a^2b/ (a^2+b^2) = 36d/25 a/ 2=3d/2 b/ 2 =4d/2 CD^2= (48d/25 -3d/2)^2-(36d/25-4d/2)^2 = (96d-75d/50)^2 + (72d-100d/50)^2 = (21d/50)^2 + (-28d/50)^2 = (441d2/2500) + (784d2/2500) = (1225d^2/2500) '''CD= 35d/50 = 7d/10 = 7d/10 = (3d+4d)/10 = (AB+AE)/10 https://edupediapublications.org/journals/index.php/IJR/article/view/3743/3589 Category:Theroem